Divide the following complex numbers. $ \dfrac{-9-7i}{-3+i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3-i}$ $ \dfrac{-9-7i}{-3+i} = \dfrac{-9-7i}{-3+i} \cdot \dfrac{{-3-i}}{{-3-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-9-7i) \cdot (-3-i)} {(-3+i) \cdot (-3-i)} = \dfrac{(-9-7i) \cdot (-3-i)} {(-3)^2 - (1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-9-7i) \cdot (-3-i)} {(-3)^2 - (1i)^2} = $ $ \dfrac{(-9-7i) \cdot (-3-i)} {9 + 1} = $ $ \dfrac{(-9-7i) \cdot (-3-i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-9-7i}) \cdot ({-3-i})} {10} = $ $ \dfrac{{-9} \cdot {(-3)} + {-7} \cdot {(-3) i} + {-9} \cdot {-1 i} + {-7} \cdot {-1 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{27 + 21i + 9i + 7 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{27 + 21i + 9i - 7} {10} = \dfrac{20 + 30i} {10} = 2+3i $